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First, let us recall the well-known qualitative test that produces a characteristic blood-red colour. The ion responsible for this colour is the ferric ion $$Fe^{3+}$$ when it reacts with the thiocyanate ion $$SCN^-$$. The reaction can be written as
$$Fe^{3+} + SCN^- \;\rightarrow\; [Fe(SCN)]^{2+}$$
The complex $$[Fe(SCN)]^{2+}$$ (and related higher thiocyanato complexes such as $$[Fe(SCN)_6]^{3-}$$) has an intense blood-red colour. Thus, whenever we see the test tube turning blood red on addition of $$SCN^-$$, the presence of $$Fe^{3+}$$ is indicated.
Now we examine each statement one by one, keeping this fact in mind.
Statement A says: “$$Fe^{2+}$$ ion also gives blood red colour with $$SCN^-$$ ion.”
But we know from the reaction above that the ferrous ion $$Fe^{2+}$$ does not form the deeply coloured thiocyanato complex. Instead, $$Fe^{2+}$$ either remains colourless/pale green or may very slowly get oxidised to $$Fe^{3+}$$ in air and then show the colour. Directly, however, $$Fe^{2+}$$ does not give the blood-red colour. Hence Statement A is false.
Statement B says: “$$Fe^{3+}$$ ion also gives blood red colour with $$SCN^-$$ ion.”
As just discussed, this is exactly the classical test for $$Fe^{3+}$$, so Statement B is true.
Statement C says: “On passing $$H_2S$$ into $$Na_2ZnO_2$$ solution, a white precipitate of $$ZnS$$ is formed.”
Let us write the ionic equation. The zincate ion present is $$ZnO_2^{2-}$$. Passing $$H_2S$$ through the alkaline solution first neutralises the basicity and then gives the sulphide:
$$ZnO_2^{2-} + 2H_2O + H_2S \;\rightarrow\; ZnS \downarrow + 4OH^-$$
The precipitate $$ZnS$$ is white. Therefore Statement C is true.
Statement D says: “Cupric ion reacts with excess of ammonia solution to give a deep blue colour of $$[Cu(NH_3)_4]^{2+}$$ ion.”
The well-known complex-formation reaction is
$$Cu^{2+} + 4NH_3 \;\rightarrow\; [Cu(NH_3)_4]^{2+}$$
The tetraamminecuprate(II) ion $$[Cu(NH_3)_4]^{2+}$$ is indeed deep blue, so Statement D is true.
Putting it all together, only Statement A is incorrect.
Hence, the correct answer is Option A.
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